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== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 10:11, 6 May 2023
Problem
Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is , where
and
are relatively prime positive integers. Find
Solution 1
Consider the three-digit arrangement, . There are
choices for
and
choices for
(since it is possible for
), and so the probability of picking the palindrome is
. Similarly, there is a
probability of picking the three-letter palindrome.
By the Principle of Inclusion-Exclusion, the total probability is
![$\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{059}$](http://latex.artofproblemsolving.com/c/a/1/ca14d4e73da3f3af21af270e9459fdb1bf0b08be.png)
Solution 2
Using complementary counting, we count all of the license plates that do not have the desired property. In order to not be a palindrome, the first and third characters of each string must be different. Therefore, there are three digit non-palindromes, and there are
three letter non palindromes. As there are
total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is
. We subtract this from 1 to get
as our probability. Therefore, our answer is
.
Solution 3
Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is thus we have
so our answer is
~Dhillonr25
Video Solution by OmegaLearn
https://youtu.be/jRZQUv4hY_k?t=98
~ pi_is_3.14
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.